Question
If Jack tosses the phone to Darnell with an initial velocity of 35 feet per second, an equation for the height h in feet of the phone after t seconds is h =-16t^2 + 35t +5.
A. If the windows is 25 feet above the ground, will Darnell have 0,1, or 2 chances to catch his phone?
B. If Darnell is unable to catch his phone, when will it hit the ground?
A. If the windows is 25 feet above the ground, will Darnell have 0,1, or 2 chances to catch his phone?
B. If Darnell is unable to catch his phone, when will it hit the ground?
Answers
GPT-4o mini
To analyze the situation of the phone being tossed to Darnell, we can break it down into the following parts.
**A. Finding how many times the height \( h \) equals 25 feet:**
We start with the equation given:
\[
h = -16t^2 + 35t + 5
\]
We set \( h \) equal to 25 feet to find the times \( t \) when the height of the phone is 25 feet:
\[
-16t^2 + 35t + 5 = 25
\]
Subtracting 25 from both sides gives:
\[
-16t^2 + 35t + 5 - 25 = 0
\]
Simplifying results in:
\[
-16t^2 + 35t - 20 = 0
\]
To make calculations easier, it's common to multiply through by -1:
\[
16t^2 - 35t + 20 = 0
\]
Next, we can apply the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 16 \), \( b = -35 \), and \( c = 20 \).
1. Calculate the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = (-35)^2 - 4(16)(20)
\]
Calculating:
\[
= 1225 - 1280 = -55
\]
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the phone will never reach a height of 25 feet.
**Conclusion for A:** Darnell will have 0 chances to catch the phone.
---
**B. Finding when the phone hits the ground:**
The phone hits the ground when \( h = 0 \). We set the height equation equal to zero:
\[
-16t^2 + 35t + 5 = 0
\]
Using the quadratic formula again, where \( a = -16 \), \( b = 35 \), and \( c = 5 \):
1. Calculate the discriminant:
\[
b^2 - 4ac = 35^2 - 4(-16)(5)
\]
Calculating:
\[
= 1225 + 320 = 1545
\]
Next, we find the values of \( t \):
\[
t = \frac{-35 \pm \sqrt{1545}}{2 \times (-16)}
\]
Calculating \( \sqrt{1545} \) gives approximately \( 39.24 \).
Now substituting back into the quadratic formula:
\[
t = \frac{-35 \pm 39.24}{-32}
\]
Calculating the two possible values for \( t \):
1. \( t = \frac{-35 + 39.24}{-32} \approx \frac{4.24}{-32} \approx -0.13 \) (not valid since time cannot be negative)
2. \( t = \frac{-35 - 39.24}{-32} = \frac{-74.24}{-32} \approx 2.32 \)
**Conclusion for B:** The phone will hit the ground approximately 2.32 seconds after being tossed.
**A. Finding how many times the height \( h \) equals 25 feet:**
We start with the equation given:
\[
h = -16t^2 + 35t + 5
\]
We set \( h \) equal to 25 feet to find the times \( t \) when the height of the phone is 25 feet:
\[
-16t^2 + 35t + 5 = 25
\]
Subtracting 25 from both sides gives:
\[
-16t^2 + 35t + 5 - 25 = 0
\]
Simplifying results in:
\[
-16t^2 + 35t - 20 = 0
\]
To make calculations easier, it's common to multiply through by -1:
\[
16t^2 - 35t + 20 = 0
\]
Next, we can apply the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 16 \), \( b = -35 \), and \( c = 20 \).
1. Calculate the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = (-35)^2 - 4(16)(20)
\]
Calculating:
\[
= 1225 - 1280 = -55
\]
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the phone will never reach a height of 25 feet.
**Conclusion for A:** Darnell will have 0 chances to catch the phone.
---
**B. Finding when the phone hits the ground:**
The phone hits the ground when \( h = 0 \). We set the height equation equal to zero:
\[
-16t^2 + 35t + 5 = 0
\]
Using the quadratic formula again, where \( a = -16 \), \( b = 35 \), and \( c = 5 \):
1. Calculate the discriminant:
\[
b^2 - 4ac = 35^2 - 4(-16)(5)
\]
Calculating:
\[
= 1225 + 320 = 1545
\]
Next, we find the values of \( t \):
\[
t = \frac{-35 \pm \sqrt{1545}}{2 \times (-16)}
\]
Calculating \( \sqrt{1545} \) gives approximately \( 39.24 \).
Now substituting back into the quadratic formula:
\[
t = \frac{-35 \pm 39.24}{-32}
\]
Calculating the two possible values for \( t \):
1. \( t = \frac{-35 + 39.24}{-32} \approx \frac{4.24}{-32} \approx -0.13 \) (not valid since time cannot be negative)
2. \( t = \frac{-35 - 39.24}{-32} = \frac{-74.24}{-32} \approx 2.32 \)
**Conclusion for B:** The phone will hit the ground approximately 2.32 seconds after being tossed.